*4.3. Repair Time*

This work uses the statistical data of time to repair failures *t*repair obtained by using the year statistics of a power supply company ("Oreolenergo") [37]. The used data that concern failures and repair time are summarized in Table 3 and shown in Figure 4.

**Table 3.** Number of failures for specific repair time.

**Figure 4.** Number of failures for specific repair time.

The values of the mathematical expectation and dispersion of the repair time were determined from this data [5,10,18]. To do this, the sampling mean *XB* was found out by Equation (11):

$$X\_B = \frac{1}{n} \times \sum\_{i}^{N\_i} (X\_i \times N\_i) \tag{11}$$
  $X\_B = \frac{0.25 \times 75 + 0.75 \times 50 + 1.25 \times 29 + 1.75 \times 38 + 2.25 \times 12 + 2.75 \times 3 + 3.25 \times 2 + 3.75 \times 3}{212} = 1$ 

where


Next, the conditional values *Ui* were determined by Equation (12):

$$\begin{array}{c} \mathcal{U}\_{i} = \frac{\mathcal{X}\_{i} - \mathcal{C}}{\frac{h}{0.25 - 0.25}}\\ \mathcal{U}\_{1} = \frac{0.25 - 0.25}{0.5} = 0 \end{array} \tag{12}$$

where


Similarly, other indicators were calculated, and the results are summarized in Table 4.


**Table 4.** Conditional values of the repair time.

The conditional sample value *UB* was determined by Equation (13):

$$\begin{aligned} lL\_B &= \frac{1}{n} \times \sum\_{i}^{N\_i} (L\_i \times N\_i) \\ lL\_B &= \frac{0 \times 75 + 1 \times 50 + 2 \times 29 + 3 \times 38 + 4 \times 12 + 5 \times 3 + 6 \times 2 + 7 \times 3}{212} = 1.5 \end{aligned} \tag{13}$$

The sample value *XB* through the conditional sample value *UB* was found by Equation (14):

$$\begin{aligned} X\_B &= \mathcal{U}\_B \times h + \mathcal{C} \\ X\_B &= 1.5 \times 0.5 + 0.25 = 1 \end{aligned} \tag{14}$$

The value of sample dispersion *DB* was determined by Equation (15):

$$D\_{B} = \frac{1}{n} \times \sum\_{i}^{N\_{i}} [(X\_{i} - X\_{B})^{2} \times N\_{i}]$$

$$D\_{B} = \frac{1}{212} \times [(0.25 - 1)^{2} \times 75 + (0.75 - 1)^{2} \times 50 + (1.25 - 1)^{2} \times 29 + (1.75 - 1)^{2} \times 38 + \dotsb \times (2.25 - 1)^{2} \times 2 + (1.75 - 1)^{2} \times 3]$$

$$(2.25 - 1)^{2} \times 12 + (2.75 - 1)^{2} \times 3 + (3.25 - 1)^{2} \times 2 + (3.75 - 1)^{2} \times 3] = 0.61$$

Root-mean-square deviation:

$$
\begin{array}{c}
\delta\_B = \sqrt{D\_B} \\
\delta\_B = \sqrt{0.61} = 0.781
\end{array}
\tag{16}
$$

The corrected root-mean-square deviation *S* was found out to obtain a more accurate value of the deviation:

$$\begin{array}{l} S = \sqrt{\frac{n}{n-1}} \times \delta\_B\\ S = \sqrt{\frac{212}{212-1}} \times 0.781 = 0.782 \end{array} \tag{17}$$

The probability of determining the repair time interval was taken to be γ = 0.95. Therefore, the value for determining the interval is *t* = 1.96.

The accuracy of the assessment:

$$\begin{aligned} \text{2F} &= \text{\textquotedblleft} = 0.95\\ F &= 0.475 \end{aligned} \tag{18}$$

The estimation deviation:

$$\frac{t \times S}{\sqrt{n}} = \frac{1.96 \times 0.782}{\sqrt{212}} = 0.105\tag{19}$$

The boundaries of the confidence interval:

$$\begin{cases} X\_B - \frac{t \times S}{\sqrt{n}} = 1 - \frac{1.96 \times 0.782}{\sqrt{212}} = 0.895 \text{ is the lower interval} \\ X\_B + \frac{t \times S}{\sqrt{n}} = 1 + \frac{1.96 \times 0.782}{\sqrt{212}} = 1.105 \text{ is the upper interval} \end{cases} \tag{20}$$

Thus, the time to repair failures was in the interval 1 ± 0.105 h with 95% probability.
